Create maximize-palindrome-length-from-subsequences.py

Author: kamyu104

Python/maximize-palindrome-length-from-subsequences.py

@@ -0,0 +1,44 @@
+# Time:  O((m + n)^2)
+# Space: O((m + n)^2)
+
+class Solution(object):
+    def longestPalindrome(self, word1, word2):
+        """
+        :type word1: str
+        :type word2: str
+        :rtype: int
+        """
+        s = word1+word2
+        dp = [[0]*len(s) for _ in xrange(len(s))]
+        result = 0
+        for j in xrange(len(s)):
+            dp[j][j] = 1
+            for i in reversed(xrange(j)):
+                if s[i] == s[j]:
+                    dp[i][j] = 2 if i+1 == j else dp[i+1][j-1] + 2
+                    if i < len(word1) <= j:
+                        result = max(result, dp[i][j])
+                else:
+                    dp[i][j] = max(dp[i+1][j], dp[i][j-1])
+        return result
+
+
+# Time:  O((m + n)^2)
+# Space: O((m + n)^2)
+class Solution2(object):
+    def longestPalindrome(self, word1, word2):
+        """
+        :type word1: str
+        :type word2: str
+        :rtype: int
+        """
+        s = word1+word2
+        dp = [[0]*len(s) for _ in xrange(len(s))]
+        for j in xrange(len(s)):
+            dp[j][j] = 1
+            for i in reversed(xrange(j)):
+                if s[i] == s[j]:
+                    dp[i][j] = 2 if i+1 == j else dp[i+1][j-1] + 2
+                else:
+                    dp[i][j] = max(dp[i+1][j], dp[i][j-1])
+        return max([dp[i][j] for i in xrange(len(word1)) for j in xrange(len(word1), len(s)) if s[i] == s[j]] or [0])